Write without summary and calculate.
$$\sum_{k=1}^n (2^k-k^2)$$
This is the part c, got the other 2 quite easily. I wrote down some of the first $n$, $1+0-1+0+7+28+79+...$
Maybe I'm asking more on how to approach this kind of a problem?
Summarizing $2^k-k^2$
1
$\begingroup$
sequences-and-series
summation
1 Answers
4
Why not just splitting it into two series?
$$\left(\color\red{\sum\limits_{k=1}^{n}2^k}\right)-\left(\color\green{\sum\limits_{k=1}^{n}k^2}\right)=\color\red{2^{n+1}-2}-\color\green{n(n+1)(2n+1)/6}$$
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0The second (green) summation is not on $k$ but on $k^2$ ! – 2017-01-05
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0@Zubzub: Oh... Thanks, you're right, I will fix that :) – 2017-01-05